A 25.00 mL suspension of milk of magnesia was added to 50.00 mL of 0.1400 M HNO3. The resulting mixture was back-titrated with 3.25 mL of 0.1040 M NAOH. The % (w/v) Mg(OH)2 (M.M. = 58.33) present in the suspension is A 1.555 15.94 0.7772 D 0.3689 B.
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- A 500ml solution of NaOH was made using 2g of NaOH(s) Three trials of titration were made with using KHP(s) as the acid dissolved with about 25ml of deionized water and 4 drops of phenolphthalein indicator. Slowly adding the NaOH solution until the clear solution had turned pink which would give us our end point and allow us to find the NaOH molarity by equivalence point. Trial 1: - 0.484g of KHP were used - initial volume of buret containing NaOH solution was 0.0 ml - final volume of buret was 23.60ml Trial 2: - 0.485g of KHP were used - initial volume was 0.0ml - final volume was 24.00ml Trial 3: - 0.486g of KHP - initial volume was 0.0ml - final volume was 23.80ml The molarity of NaOH was found by using the moles of KHP(as at equivalence, both solutions are balanced in moles) divided by the total volume of NaOH used to neutralize the solution. Giving us 0.100M for trial 1, 0.099M for trial 2, and 0.100M for trial 3. Make a rough sketch of a titration curve that…A 2.054 g of macrogol monostearate (average molecular weight 706. 5) was added to a 200 ml flask and 25 ml of an ethanolic solution of potassium hydroxide (molecular weight 56.1, ca 0.5 M) was added. The sample was heated under a reflux condenser for 1 hour. The excess of alkali was then titrated with 0.5016 M hydrochloric acid using phenolphthalein solution as an indicator. The operation was repeated without the macrogol monostearate. Results Volume of HCI required to titrate the excess alkali = 18.35 ml Volume of HCI required to titrate the blank = 24.03 ml Calculate the saponification value for the macrogol stearate. Answer: Blank mg/g20 aspirin tablets labeled 80mg were dissolved in 100mL of 90% ethanol. A 10mL aliquot was taken and was used for assay. The analyte followed usual process and was treated with 50mL of 0.1000N NaH and was titrated with 35mL 0.1050N H2O4 until the solution achieved completion. Calculate the % content of the total aspirin capsules and the actual label claim
- Calcium fluoride is considered as a relatively insoluble compound and therefore lime or slakedlime has been considered as a possible material to remove excess fluoride in water of boreholesin certain parts of the country. The solubility product of CaF2 is Ksp = 3 x 10 – 11 and that ofCa(OH)2 isKsp =8x10-61. How much lime can be added to the water to remove 10 mg of F- ion per litre ofborehole water?(The atomic masses are Ca: 40.08; F: 19.00; O: 16; H: 1)1. A 1.2-gram sample of lanolin was treated with Wij’s solution and excess potassium iodide solution. The liberated iodine reacted with 30 ml of 0.1 N sodium thiosulfate solution. If the iodine value was determined as 12.69, what is the volume used in blank titration? 2. A fat sample with combination of acids contain standard hydrochloric acid for blank and sample with 8mL and 5mL respectively. The normality of the standard hydrochloric acid is 0.93N and the weight of the sample is 3 grams. Calculate the saponification value. 3. A 3.50-gram sample of Streptomycin powder was tested for its water content. If the water equivalence factor of the KF reagent was 4.6, what is the percentage water content of the sample if 9.2 ml of the KF reagent was used? 4. A 500mg oil sample is taken from a conical flask and is dissolved in 50mL distilled alcohol. An indicator is added and is then titrated against 0.112N KOH until a slight pink color appears. It took 17.6mL of the titrant to reach the…Dissolved 0.273 grams of pure sodium oxalate (Na2C204) in distilled water and added sulfuric acid and titration the solution at 70 °C by using 42.68 ml of KMNO4 solution and has exceeded end point limits by using 1.46 ml of standard oxalic acid (H2 C204) with 0.1024 N. Calculate the normlity of KMNO4. Note that the molecular weight of sodium oxalate (Na2C204) = 134 and its equivalent weight = 67 * %3D
- A STOCK SOLUTION containing 0.1581 g/L K2CrO4 was prepared.In order to make the CALIBRATION STANDARD, 5 ml of the STOCK was transferredinto a 50ml volumetric flask and then diluted with an appropriate solvent.Calculate:(a) The ppm of K2CrO4 in the CALIBRATION STANDARD.(b) The molarity of K2CrO4 in the CALIBRATION STANDARD. (c) Calculate the molar absorptivity of K2CrO4 (at 371.0 nm). Assume that Beer's Law isobeyed over this concentration range.At 371.0 nm, this CALIBRATION STANDARD in a cell of path length 1.00 cm gave a %T of 59.752.A 0.1093-g sample of impure Na2CO3( Molecular weight 106) was analyzed by the Volhard method. After adding 50.00 mL of 0.06911 M AGN03 (Molecular weight 169.87), the sample was back titrated with 0.05781 M KSCN, requiring 27.36 mL to reach the end point. Report the purity of the Na2CO3 sample.The molar solubility of MgCO3 (Ksp = 3.50 x 10-8) in distilled water at room temperature is ______ M. Titration of a 50.00-mL aliquot of the saturated solution will require ______mL of 0.005000 M HCl to reach the phenolphthalein endpoint.
- 0.1 g of the mixture of na2so4 and k2so4 is taken and 100 ml of solution is prepared. 10 ml of this prepared solution is placed in a beaker and some distilled water is added. A mass of 15.5 mg is obtained by precipitation with Bacl2 at PH=5, then filtering and bringing to a constant weight at 800 °C. Calculate the percentages of Na2so4 and K2so4 in the mixture accordingly.Dissolved 0.273 grams of pure sodium oxalate (NaCO) in distilled water and added sulfuric acid and titration the solution at 70 ° C by using 42.68 ml of KMNO. solution and has exceeded end point limits by using 1.46 ml of standard oxalic acid (HCO) with 0.1024 N Calculate the normlity of KMNO Note that the molecular weight of sodium oxalate (NaCO) = 134 and its equivalent weight = 67A 0.1093-g sample of impure Na2CO3 was analyzed by the Volhard method. After adding 50.00 mL of 0.06911 M AgNO3, the sample was back-titrated with 0.05781 M KSCN, requiring 27.36 mL to reach the endpoint. Report the purity of the Na2CO3 sample. [Ans. 90.9 % (w /w )]
